5.6. S Gate and S\(^\dagger\) Gate

We shall call \(S\) and \(S^\dagger\) gates SGate and SdgGate, respectively.

API References: SGate
API References: SdgGate

Definition

\(S |0\rangle = |0\rangle, \qquad S |1\rangle = i|1\rangle\)
\(S^\dagger |0\rangle = |0\rangle, \qquad S^\dagger |1\rangle = -i|1\rangle\)

The qiskit circuit symbol is \(S\) for SGate and \(Sdg\) for SdgGate. They appear in quantum circuits as

from qiskit import QuantumCircuit
qc=QuantumCircuit(1)
qc.s(0)
qc.sdg(0)
qc.draw()

   ┌───┐┌─────┐
q: ┤ S ├┤ Sdg ├
   └───┘└─────┘

Acting on a superposition state

When SGate and SdgGate are applied to a super position state the coefficient is swapped. That is

(5.8)\[ S \left (c_0 |0\rangle + c_1 |1\rangle\right) = c_0 |0\rangle + i c_1 |1\rangle \]

(5.9)\[ S^\dagger \left (c_0 |0\rangle + c_1 |1\rangle\right) = c_0 |0\rangle - i c_1 |1\rangle \]

Notice that the relative phase changes by \(e^{\pm i\pi/2} = \pm i\). Recall that \(Z\) changes the relative phase by \(e^{i \pi} = -1\). Further notice that \((e^{\pm i \pi/2})^2 = e^{i \pi}\). Hence, \(S^2 = (S^\dagger)^2 = Z\). Because of this relation, \(S\) is sometimes expressed as \(\sqrt{Z}\).

Setting \(c_0=c_1=\frac{1}{\sqrt{2}}\), we find basis transformation

(5.10)\[ S|+\rangle = |L\rangle, \quad S|-\rangle = |R\rangle \]

(5.11)\[ S^\dagger|+\rangle = |R\rangle, \quad S^\dagger|-\rangle = |L\rangle \]

Since \(S\) is unitary \(S S^\dagger = I\) and thus \(S^{-1}=S^\dagger\) and \((S^\dagger)^{-1} = S\). Now the inverse of (5.10) and SdgGate- fwd are

(5.12)\[ S^\dagger |L\rangle = |+\rangle, \quad S^\dagger|R\rangle = |-\rangle \]

(5.13)\[ S|L\rangle = |-\rangle, \quad S|R\rangle = |+\rangle \]

Combining HGate and SGate, we can transform the computational basis \(\{|0\rangle, |1\rangle\}\) to \(\{|L\rangle, |R\rangle\}\) by \(S H\) and its inverse is \(H S^\dagger\). Now, we know we can move from one basis to another by \(H\), \(S\), and \(SH\).

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Qiskit Example 5.6.1  We demonstrate the above basis set transformation using Qiskit. First, we construct a quantum circuit corresponding to the following transformation

\[ |0\rangle \xrightarrow{H} |+\rangle \xrightarrow{S} |L\rangle \xrightarrow{S^\dagger} |+\rangle \xrightarrow{H} |0\rangle \]

Notice that the whole operation can be written as \(H S^\dagger S H |0\rangle\) it can be simplified as

\[ H S^\dagger S H |0\rangle = H (S^\dagger S) H = H I H = H^2 = I \]

Hence, the whole operation does nothing at all. In order to avoid unnecessary computation like this, we need to understand the properties of gates. You will surprise that a long circuit can be significantly shortened by contracting gates.

%%capture
from qiskit import *
from qiskit.visualization import visualize_transition

qc=QuantumCircuit(1)

qc.h(0)
qc.s(0)
qc.sdg(0)
qc.h(0)

movie=visualize_transition(qc,fpg=50, spg=1)

qc.draw('mpl')

movie

Once Loop Reflect

Additional useful Properties

\(S\, S = S^\dagger \, S^\dagger = Z\) implies that

1. \(S = Z S^\dagger = S^\dagger Z, \quad S^\dagger = Z S = S Z\).

2. \(Z = S Z S^\dagger = S^\dagger Z S\)

3. \(S = Z S Z, \quad S^\dagger = Z S^\dagger Z\).

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Actions on X, Y, and Z-basis